If 3.50g of Na3Po4 and 6.40g of Ba(No3)2 are added together how many grams of Ba3(PO4)2 will be produced? Which is the limiting reactant?
Please explains step by step
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1. Write the equation and balance it.
2Na3PO4 + 3Ba(NO3)2 ==> Ba3(PO4)2 + 6NaNO3
2. Convert 3.50 g Na3PO4 to moles. moles = grams/molar mass.
3. Convert 6.40 g Ba(NO3)2 to moles. Same process.
4a. Using the coefficients in the balanced equation, convert mole Na3PO4 to moles of the product.
4b. Same process, convert moles Ba(NO3)2 to moles of the product.
4c. It is quite likely that the answers in 4a and 4b will not agree which means one of them is wrong. The correct answer, in limiting reagent problems, is ALWAYS the smaller one and the reagent providing that answer is the limiting reagent.
5. Using the smaller value, convert moles of the product to grams. g = moles x molar mass.
Calculation
To determine the grams of Ba3(PO4)2 produced when 3.50g of Na3PO4 and 6.40g of Ba(NO3)2 are added together, we need to use stoichiometry and the concept of limiting reactants.
Step 1: Write the balanced chemical equation:
2 Na3PO4 + 3 Ba(NO3)2 -> Ba3(PO4)2 + 6 NaNO3
Step 2: Calculate the molar mass of Na3PO4:
Na: 22.99 g/mol x 3 = 68.97 g/mol
P: 30.97 g/mol
O: 16.00 g/mol x 4 = 64.00 g/mol
Total molar mass of Na3PO4 = 68.97 + 30.97 + 64.00 = 163.94 g/mol
Step 3: Calculate the number of moles of Na3PO4:
moles = mass / molar mass = 3.50 g / 163.94 g/mol = 0.02134 mol
Step 4: Calculate the molar mass of Ba(NO3)2:
Ba: 137.33 g/mol
N: 14.01 g/mol x 2 = 28.02 g/mol
O: 16.00 g/mol x 6 = 96.00 g/mol
Total molar mass of Ba(NO3)2 = 137.33 + 28.02 + 96.00 = 261.35 g/mol
Step 5: Calculate the number of moles of Ba(NO3)2:
moles = mass / molar mass = 6.40 g / 261.35 g/mol = 0.0245 mol
Step 6: Use the stoichiometry of the balanced equation to determine the moles of Ba3(PO4)2 that can be produced:
From the balanced equation, we can see that 3 moles of Ba(NO3)2 reacts with 1 mole of Ba3(PO4)2.
Based on this ratio, we can calculate the moles of Ba3(PO4)2 that can be produced:
moles of Ba3(PO4)2 = (0.0245 mol Ba(NO3)2) / (3 mol Ba(NO3)2) x (1 mol Ba3(PO4)2) = 0.008167 mol Ba3(PO4)2
Step 7: Calculate the theoretical mass of Ba3(PO4)2:
mass = moles x molar mass = 0.008167 mol Ba3(PO4)2 x molar mass of Ba3(PO4)2
The molar mass of Ba3(PO4)2 is calculated by adding together the molar masses of each element:
Ba: 137.33 g/mol x 3 = 411.99 g/mol
P: 30.97 g/mol
O: 16.00 g/mol x 8 = 128.00 g/mol
Total molar mass of Ba3(PO4)2 = 411.99 + 30.97 + 128.00 = 570.96 g/mol
mass = 0.008167 mol Ba3(PO4)2 x 570.96 g/mol = 4.657 g Ba3(PO4)2
Therefore, when 3.50g of Na3PO4 and 6.40g of Ba(NO3)2 are added together, approximately 4.657g of Ba3(PO4)2 will be produced.
Step 8: Determine the limiting reactant:
The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed. To find the limiting reactant, we compare the moles of each reactant to the stoichiometry of the balanced equation.
From step 3, the moles of Na3PO4 is 0.02134 mol.
From step 5, the moles of Ba(NO3)2 is 0.0245 mol.
Based on the balanced equation, we can see that 3 moles of Ba(NO3)2 react with 2 moles of Na3PO4. Therefore, the stoichiometric ratio of Na3PO4 to Ba(NO3)2 is 2:3.
To find the limiting reactant, we compare the moles of Na3PO4 and Ba(NO3)2 to the stoichiometric ratio:
Moles of Na3PO4 (0.02134 mol) / 2 = 0.01067 mol
Moles of Ba(NO3)2 (0.0245 mol) / 3 = 0.008167 mol
Comparing these values, we can see that 0.008167 mol of Ba(NO3)2 is less than 0.01067 mol of Na3PO4. Therefore, the limiting reactant is Ba(NO3)2.
I hope this step-by-step explanation helps!
To find out how many grams of Ba3(PO4)2 will be produced when 3.50g of Na3PO4 and 6.40g of Ba(NO3)2 are mixed, we need to follow these steps:
Step 1: Write the balanced chemical equation.
The balanced equation for the reaction between Na3PO4 and Ba(NO3)2 is:
3Na3PO4 + 2Ba(NO3)2 → Ba3(PO4)2 + 6NaNO3
Step 2: Calculate the number of moles of each reactant.
To calculate the number of moles, you need to divide the mass of each reactant by its molar mass. The molar mass can be calculated by adding up the atomic masses of the elements in each compound.
The molar mass of Na3PO4 is:
(3 x molar mass of Na) + molar mass of P + (4 x molar mass of O)
= (3 x 22.99 g/mol) + 30.97 g/mol + (4 x 16.00 g/mol)
= 163.94 g/mol
To calculate the number of moles of Na3PO4, divide the mass (3.50g) by its molar mass (163.94 g/mol):
moles of Na3PO4 = 3.50g / 163.94 g/mol
The molar mass of Ba(NO3)2 is:
molar mass of Ba + (2 x molar mass of N) + (6 x molar mass of O)
= 137.33 g/mol + (2 x 14.01 g/mol) + (6 x 16.00 g/mol)
= 261.35 g/mol
To calculate the number of moles of Ba(NO3)2, divide the mass (6.40g) by its molar mass (261.35 g/mol):
moles of Ba(NO3)2 = 6.40g / 261.35 g/mol
Step 3: Determine the limiting reactant.
The limiting reactant is the reactant that is completely consumed first and limits the amount of product formed. To determine the limiting reactant, calculate the moles of the product formed from each reactant and compare them.
Using the balanced equation, we see that the mole ratio between Na3PO4 and Ba3(PO4)2 is 3:1. This means that for every 3 moles of Na3PO4 reacted, 1 mole of Ba3(PO4)2 is formed.
Converting the moles of Na3PO4 to moles of Ba3(PO4)2:
moles of Ba3(PO4)2 formed from Na3PO4 = (moles of Na3PO4) x (1 mole Ba3(PO4)2 / 3 moles Na3PO4)
Converting the moles of Ba(NO3)2 to moles of Ba3(PO4)2:
moles of Ba3(PO4)2 formed from Ba(NO3)2 = (moles of Ba(NO3)2) x (1 mole Ba3(PO4)2 / 2 moles Ba(NO3)2)
Step 4: Compare the moles of Ba3(PO4)2 formed from each reactant.
Compare the moles calculated in step 3 for each reactant. The reactant that produces the smaller number of moles of Ba3(PO4)2 is the limiting reactant.
Step 5: Calculate the mass of Ba3(PO4)2 formed.
To calculate the mass of Ba3(PO4)2 formed, multiply the number of moles of Ba3(PO4)2 by its molar mass.
Step 6: Determine the excess reactant.
The excess reactant is the reactant that is not completely consumed and is left over after the reaction is complete. Calculate the moles of the excess reactant used and subtract it from the initial moles of that reactant to find the moles remaining.
By following these steps, you will be able to find the grams of Ba3(PO4)2 produced and determine the limiting reactant in the given reaction.